Solutions Manual: Ch14-PMTS-s: Review Questions [PDF]

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Solutions Manual: Ch14-PMTS-s Review Questions 14.1 Define predetermined motion time system. Answer: A predetermined motion time system is a database of basic motion elements and their associated normal time values, together with a set of procedures for applying the data to analyze manual tasks and establish standard times for the tasks. 14.2 What are the steps in applying a predetermined motion time system? Answer: The steps listed in the text are the following: (1) Synthesize the method that would be used by a worker to perform the task (or analyze the method that is being used by a worker in an existing task). The method is described in terms of the basic motion elements comprising the task, based on a defined workplace layout and set of tools (if tools are used). (2) Retrieve the normal time value for each motion element, based on the work variables and conditions under which the element will be (is) performed. Sum the normal times for all motion elements to determine the normal time for the task. (3) Evaluate the method to make improvements by eliminating motions, reducing distances, and so on. The objective of this evaluation is to reduce the normal time. (4) Apply allowances to the normal time to determine the standard time for the task. 14.3 What is the difference between a first level PMTS and a higher-level PMTS? Answer: First level systems use the basic motion elements (e.g., reach, grasp), while higher level systems combine several basic motion elements into motion sequences. First level PMT systems tend to be very detailed, with body motions differentiated very precisely in their databases. Higher level systems use condensed databases, with fewer body motions contained in the tables and longer time values for each motion sequence. 14.4 What is a motion aggregate in a higher-level PMTS? Answer: A motion aggregate is a combination of several basic motion elements performed together in a sequence. An example is “get an object,” which consists of the basic motions reach and grasp. 14.5 What are the advantages of a higher-level PMTS compared to a first level PMTS? Answer: Advantages of a higher-level PMTS include faster application speed, longer cycle times are feasible, and easier to apply. 14.6 What is the unit of time used in Methods-Time Measurement? Answer: The unit of time in MTM is the TMU or time measurement unit, which = 0.00001 hr or 0.036 sec. 14.7 What does the acronym MOST stand for? Answer: Maynard Operation Sequence Technique. 14.8 What is the primary focus of MOST in terms of type of work activity? Answer: The primary focus of MOST is on the movement of objects, such as parts and tools from one location to another in the workplace. 1 Work Systems and the Methods, Measurement, and Management of Work by Mikell P. Groover. ISBN 0-13-140650-7. © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

Solutions Manual: Ch14-PMTS-s 14.9 What are the motion aggregates in MOST called? Answer: Activity sequence models. 14.10 What is the difference between General Move and Controlled Move in MOST? Answer: The General Move is applicable when an object is moved freely through the air from one location to another. The Controlled Move is used when an object is moved through a path that is somehow constrained, such as when an object is slid across a surface or when the object is attached to something else and it can only be moved through a controlled path.

Problems MTM-1 14.1 A worker seated at a table performs a REACH. The sought-after object is jumbled with other objects in a tote pan, and the distance of the reach is 18 in. Determine the MTM-1 symbol and normal time in TMUs for this motion element. Solution: MTM-1 symbol = R18C, Tn = 18.4 TMU 14.2 An assembly worker standing at a workbench performs a MOVE. The object being moved weighs 10 lb. It is moved to an exact location a distance of 20 in. Determine the MTM-1 symbol and normal time in TMUs for this motion element. Solution: MTM-1 symbol = M20C10, Tn = 3.9 + 1.11(22.1) = 3.9 + 24.5 = 28.4 TMU 14.3 An office worker sitting at a desk performs a MOVE. The moved object is a file weighing less than 1 lb. The distance is an approximate location 14 in. away from the starting point. Determine the MTM-1 symbol and normal time in TMUs for this motion element. Solution: MTM-1 symbol = M14B1, Tn = 0 + 14.6 = 14.6 TMU 14.4 A mechanic performs two POSITION elements in sequence. The first requires an alignment along longitudinal axes of the objects, and the second requires an orientation of their rotational axes. Both elements can be classified as close fits with no symmetry, and the objects are easy to handle. (a) Determine the MTM-1 symbols and normal times in TMUs for these motion elements. (b) What is the total time for both elements in sec? Solution: (a) MTM-1 symbol = P2NSE, Tn = 21.0 TMU for one POSITION (b) Tn = 2 x 21.0 = 42.0 TMU. In seconds, Tn = 42(0.036) = 1.51 sec 14.5 What is the MTM-1 normal time in TMUs that should be allowed an operator who must read an instruction set before proceeding to perform a processing task, in which the instruction document contains 150 words? How many seconds is this? Solution: Reading Tn = 5.05(150) = 757.5 TMU. In seconds, Tn = 757.5(0.036) = 27.25 sec 14.6 In performing a certain manual task, a worker must walk a distance of 30 ft (one way) as one of the elements. (a) What is the MTM-1 normal time in TMUs that should be allowed for the element? (b) What is this walking time in sec? Solution: (a) Walking MTM-1 symbol = W-FT, Tn = 30(5.3) = 159 TMU 2 Work Systems and the Methods, Measurement, and Management of Work by Mikell P. Groover. ISBN 0-13-140650-7. © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

Solutions Manual: Ch14-PMTS-s (b) In seconds, Tn = 159(0.036) = 5.72 sec 14.7 For safety reasons, a worker’s eye travel time in a certain operation must be separated from the manual elements that follow. The distance the worker’s eyes must travel is 20 in. The perpendicular distance from her eyes to the line of travel is 2 ft. No refocus is required. What is the MTM-1 normal time in TMUs that should be allowed for the eye travel element? Solution: T = 20 in, D = 2(12) = 24 in Tn = 15.2(20/24) = 12.7 TMU 14.8 A work element in a manual assembly task consists of the following MTM-1 elements: (1) R16C, (2) G4A, (3) M10B5, (4) RL1, (5) R14B, (6) G1B, (7) M8C3, (8) P1NSE, and (9) RL1. (a) Determine the normal times in TMUs for these motion elements. (b) What is the total time for this work element in sec? Solution: (a) The individual motion element times are given as follows: (1) R16C, Tn = 17.0 TMU; (2) G4A, Tn = 7.3 TMU; (3) M10B5, Tn = 2.2 + 1.06(12.2) = 15.1 TMU; (4) Rl1, Tn = 2.0 TMU; (5) R14B, Tn = 14.4 TMU; (6) G1B, Tn = 3.5 TMU; (7) M8C3, Tn = 2.2 + 1.06(11.8) = 14.7 TMU; (8) P1NSE, Tn = 10.4 TMU; and (9) RL1, Tn = 2.0 TMU. Total = 86.4 TMU. (b) In seconds, Tn = 86.4(0.036) = 3.1 sec 14.9 A work element in a machine maintenance operation consists of the following MTM-1 elements: (1) W5P, i.e., walk 5 paces, (2) B, i.e., bend (3) R14B, (4) G1A, (5) AB, i.e., arise from bend, (6) W5P, (6) M12B2, (7) P1SSE, (8) RL1. (a) Determine the normal times in TMUs for these motion elements. (b) What is the total time for this work element in sec? Solution: (a) The individual motion element times are given as follows: (1) W5P, Tn = 5(15.0) = 75.0 TMU; (2) B, Tn = 29.3 TMU; (3) R14B, Tn = 14.4 TMU; (4) G1A, Tn = 2.0 TMU; (5) AB, Tn = 31.9 TMU; (6) W5P, Tn = 5(15.0) = 75.0 TMU; (7) P1SSE, Tn = 9.1 TMU; and (8) RL1, Tn = 2.0 TMU. Total = 251.8 TMU. (b) In seconds, Tn = 251.8(0.036) = 9.1 sec

MOST 14.10 Develop the activity sequence model and determine the normal time for the following work activity: A worker walks three steps, picks up a screwdriver from the floor, returns to his original position, and places the screwdriver on his worktable. Solution: Activity sequence model = A6B6G1A6B0P1A0 Tn = 10(6 + 6 + 1 + 6 + 1) = 10(20) = 200 TMU (7.2 sec) 14.11 Develop the activity sequence model and determine the normal time for the following work activity: A clerk walks 8 steps, bends and picks up a file folder from the floor, places it on the counter within reach at that location, and then returns to her original location. Solution: Activity sequence model = A16B6G1A1B0P1A16 Tn = 10(16 + 6 + 1 + 1 + 1 + 16) = 10(41) = 410 TMU (14.7 sec) 3 Work Systems and the Methods, Measurement, and Management of Work by Mikell P. Groover. ISBN 0-13-140650-7. © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

Solutions Manual: Ch14-PMTS-s 14.12 Develop the activity sequence model or models and determine the normal time for the following work activity: An assembly worker obtains four bolts in one hand from a bin located 10 in away on his worktable and puts one bolt each into four holes in the bracket within easy reach in front of him. Solution: Activity sequence model: A1B0G3A1B0P1A0 Tn = 6 TMU A0B0G0A1B0P1A0 Tn = 2 TMU A0B0G0A1B0P1A0 Tn = 2 TMU A0B0G0A1B0P1A0 Tn = 2 TMU Total Tn = 12 TMU x 10 = 120 TMU (4.3 sec) 14.13 Develop the activity sequence model and determine the normal time for the following work activity: A machinist standing in front of his milling machine, grasps the waist level feed lever on the machine, and rotates the lever one crank to engage the feed. The process time to mill the part is 50 sec. There is no alignment and no action by the worker at the end of the process time. Solution: Process time in TMU = 50(27.8) = 1390 TMU Activity sequence model = A1B0G1M3X139I0A0 Tn = 10(1 + 1 + 3 + 139) = 10(144) = 1440 TMU (51.8 sec) 14.14 Develop the activity sequence model and determine the normal time for the following work activity: A material handling worker grasps a carton weighing 20 lb on a counter and slides it along the countertop a distance of 2 ft. Solution: Activity sequence model = A1B0G3M6X0I0A0 Tn = 10(1 + 3 + 6) = 10(10) = 100 TMU (3.6 sec) 14.15 Develop the activity sequence model and determine the normal time for the following work activity: A drill press operator reaches 20 cm (8 in) and pulls the feed lever down to engage the feed motion, which takes 12 sec. Solution: Process time in TMU = 12(27.8) = 333.6 TMU Activity sequence model = A1B0G1M1X33I0A0 Tn = 10(1 + 1 + 1 + 33) = 10(36) = 360 TMU (12.9 sec) 14.16 Develop the activity sequence model and determine the normal time for the following work activity: A worker walks three steps, picks up a screw from his worktable, walks back to his initial location, positions the screw into a threaded hole, and turns it five spins with his fingers. Solution: Activity sequence model = A6B0G1A6B0P3F8A0B0P0A0 Tn = 10(6 + 1 + 6 + 3 + 8) = 10(24) = 240 TMU (8.6 sec) 14.17 Develop the activity sequence model and determine the normal time for the following work activity: A worker picks up a screwdriver within reach from his worktable, positions it onto the head of a screw, fastens the screw with six turns, and lays the screwdriver aside. Solution: Activity sequence model = A1B0G1A1B0P3F16A1B0P1A0 Tn = 10(1 + 1 + 1 + 3 + 16 + 1 + 1) = 10(24) = 240 TMU (8.6 sec) 4 Work Systems and the Methods, Measurement, and Management of Work by Mikell P. Groover. ISBN 0-13-140650-7. © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

Solutions Manual: Ch14-PMTS-s 14.18 Develop the activity sequence model and determine the normal time for the following work activity: An assembly worker on a production line obtains an Allen key within reach, positions it 15 cm (6 in) onto a bolt head, cranks it 7 times to seat the bolt, and then sets the key aside. Solution: Activity sequence model = A1B0G1A0B0P3F16A1B0P1A0 Tn = 10(1 + 1 + 3 + 16 + 1 + 1) = 10(23) = 230 TMU (8.3 sec) 14.19 Express the MTM-1 motion elements in Problem 14.8 as one or more MOST activity sequence models with index numbers. (a) Determine the normal times in TMUs for these sequence activity models. (b) What is the total time for this (these) sequence activity model(s) in sec? (c) How do the MOST normal times compare with the normal times from MTM-1? Solution: For the MTM-1 sequence R16C, G4A, M10B5, and RL1, the MOST activity sequence model = A1B0G3A0B0P1A0, and Tn = 10(1 + 3 + 1) = 50 TMU For the MTM-1 sequence R14B, G1B, M8C3, P1NSE, and RL1, the MOST activity sequence model = A1B0G1A0B0P1A0, and Tn = 10(1 + 1 + 1) = 30 TMU (b) Tn for the complete sequence = 50 + 30 = 80 TMU (2.88 sec) (c) The corresponding MTM-1 time was 86.4 TMU (3.11 sec). This is a difference of 6.4 TMU (0.23 sec) or 7.4% based on the MTM-1 value as the base. 14.20 Express the MTM-1 motion elements in Problem 14.9 as one or more MOST activity sequence models with index numbers. (a) Determine the normal times in TMUs for these sequence activity models. (b) What is the total time for this (these) sequence activity model(s) in sec? (c) How do the MOST normal times compare with the normal times from MTM-1? Solution: For the MTM-1 sequence W5P, B, R14B, G1A, AB, W5P, M12B2, P1SSE, and RL1, the MOST activity sequence model = A10B6G1A10B0P1A0, (b) Tn = 10(10 + 6 + 1 + 10 + 1) = 10(28) = 280 TMU (10.08 sec) (c) The corresponding MTM-1 time was 251.8 TMU (9.06 sec). This is a difference of 28.2 TMU (1.02 sec) or 11.2% based on the MTM-1 value as the base. 14.21 A work element in a worker-machine cycle has been reduced to the following two MOST activity sequence models: A3B10G3A6B10P3A3 A3B0G1M1XTI0A3 In the second sequence model, the process time is known to be 9.3 seconds. This value must be converted to the correct index value, symbolized by the subscript T for the X parameter in the second sequence model. (a) What is the correct value of T? (b) Determine the normal time in seconds that would be allocated for this work element. Conversion factor = 27.8 TMU per second. Solution: (a) For the process time = 9.3 sec, the index value for the X parameter = 9.3(27.8/10) = 25.85 rounded to 26. The value of T = 26. 5 Work Systems and the Methods, Measurement, and Management of Work by Mikell P. Groover. ISBN 0-13-140650-7. © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.

Solutions Manual: Ch14-PMTS-s (b) Tn = 10(3 + 10 + 3 + 6 + 10 + 3 + 3) + 10(3 + 0 + 1 + 1 + 26 + 0 + 3) = 720 TMU In seconds, Tn = 720(0.036) = 25.9 sec

Miscellaneous 14.22 The normal time for walking in MTM-1 is 5.3 TMU per ft distance. (a) Using this value determine the amount of time in min that would be required to walk one mile (5280 ft). (b) How does this compare with the traditional benchmark of standard performance of walking at 3 miles/hr? Solution: (a) In MTM-1, Tn for walking = 5.3 TMU per ft distance For one mile, Tn = 5280(5.3) = 27,984 TMU In minutes, Tn = 27,984(0.036/60) = 16.79 min (b) The speed of walking a mile in 16.79 min = 60/16.79 = 3.57 mi/hr. This is about 19% faster than the traditional benchmark of standard performance or 3.0 mi/hr. 14.23 The normal time for walking in MOST is 6.9 TMU per ft distance.1 (a) Using this value determine the amount of time in min that would be required to walk one mile (5280 ft). (b) How does this compare with the traditional benchmark of standard performance of walking at 3 miles/hr? Solution: (a) In MOST, Tn for walking = 6.9 TMU per ft distance For one mile, Tn = 5280(6.9) = 36,432 TMU In minutes, Tn = 36,432(0.036/60) = 21.85 min (b) The speed of walking a mile in 21.85 min = 60/21.85 = 2.75 mi/hr. This is about 8.3% slower than the traditional benchmark of standard performance or 3.0 mi/hr.

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Based on tabulated values in [Zandin, 1990, p 35]. 6 Work Systems and the Methods, Measurement, and Management of Work by Mikell P. Groover. ISBN 0-13-140650-7. © 2007 Pearson Education, Inc., Upper Saddle River, NJ. All rights reserved.